A small ball (mass m) falling under gravity in a viscous medium experience a drag force proportional to the instantaneous speed u such that `F_(drag)=ku`. Then the terminal speed of ball within viscous medium is

Options

A. `k/(mg)`
B. `(mg)/k`
C. `sqrt((mg)/(k))`
D. None of these

Explanation

To find the terminal speed of a small ball falling under gravity in a viscous medium, we can follow these steps:

### Step 1: Identify the Forces Acting on the Ball
When the ball is falling, there are three main forces acting on it:
1. The weight of the ball (downwards), which is given by \( F_{\text{weight}} = mg \).
2. The drag force (upwards), which is proportional to the instantaneous speed \( u \), given by \( F_{\text{drag}} = ku \).
3. The buoyant force (upwards), which is equal to the weight of the fluid displaced by the ball, given by \( F_{\text{buoyancy}} = V \rho_l g \), where \( V \) is the volume of the ball, \( \rho_l \) is the density of the liquid, and \( g \) is the acceleration due to gravity.

### Step 2: Write the Equation of Motion
At terminal velocity, the net force acting on the ball is zero. Therefore, we can set up the equation:
\[ F_{\text{weight}} = F_{\text{drag}} + F_{\text{buoyancy}} \]
Substituting the expressions for these forces, we have:
\[ mg = ku + V \rho_l g \]

### Step 3: Express Volume of the Ball
The volume \( V \) of the ball can be expressed in terms of its density \( \rho_b \) and mass \( m \):
\[ V = \frac{m}{\rho_b} \]
Substituting this into the equation gives:
\[ mg = ku + \left(\frac{m}{\rho_b}\right) \rho_l g \]

### Step 4: Rearrange the Equation
Rearranging the equation to isolate \( u \) (the terminal speed) gives:
\[ mg - \left(\frac{m}{\rho_b}\right) \rho_l g = ku \]
Factoring out \( g \) from the left side:
\[ g \left( m - \frac{m \rho_l}{\rho_b} \right) = ku \]

### Step 5: Solve for Terminal Velocity
Now, we can solve for the terminal velocity \( u \):
\[ u = \frac{g \left( m - \frac{m \rho_l}{\rho_b} \right)}{k} \]
This simplifies to:
\[ u = \frac{mg \left( \rho_b - \rho_l \right)}{k \rho_b} \]
Thus, the terminal speed \( v_t \) can be expressed as:
\[ v_t = \frac{g (\rho_b - \rho_l)}{k} \]

### Final Answer
The terminal speed of the ball within the viscous medium is:
\[ v_t = \frac{g (\rho_b - \rho_l)}{k} \]

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